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At what height from the surface of the earth, the acceleration due to gravity will be ${\frac{1}{4}}^{th}$ of g on earth. R – Radius of earth

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3 R

2 R

answer is C.

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Detailed Solution

$g^{l} = g {(\frac{R}{R + h})}^{2} \Rightarrow \frac{g}{4} = g (\frac{R^{2}}{R + h})$
$\begin{array}{l} ∴ \frac{R}{R + h} = \frac{1}{2} \\ 2 R = R + h \\ h = R \end{array}$

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